# Convolution of Two Monomials

1. May 25, 2010

### Char. Limit

So I wanted to find the convolution of two monomials (don't ask me why) in the most general way I knew how. Of course, I only knew how for positive integer powers, but maybe someone can tell me how to do it for positive real powers (Gamma function maybe?). Anyway, here's what I did, replacing tau with s to make it easier to write:

$$t^a \star t^b = \int_0^t s^a \left(t-s\right)^b ds$$

$$= \int_0^t s^a \left(t^b + \left(\left(-1\right)^1 b\right) s t^{b-1} + \frac{\left(-1\right)^2 b \left(b-1\right)}{2} s^2 t^{b-2} + ... + \frac{\left(-1\right)^{b-2} b \left(b-1\right)}{2} s^{b-2} t^2 + \left(\left(-1\right)^{b-1} b\right) t s^{b-1} + \left(-1\right)^b s^b\right) ds$$

$$= \int_0^t t^b s^a + \left(\left(-1\right)^1 b\right) t^{b-1} s^{a+1} + \frac{\left(-1\right)^2 b \left(b-1\right)}{2} t^{b-2} s^{a+2} + ... + \frac{\left(-1\right)^{b-2} b \left(b-1\right)}{2} t^2 s^{a+b-2} + \left(\left(-1\right)^{b-1} b\right) t s^{a+b-1} + \left(-1\right)^b s^{a+b} ds$$

$$= \frac{t^b s^{a+1}}{a+1} + \frac{\left(-1\right)^1 b t^{b-1} s^{a+2}}{a+2} + \frac{\left(-1\right)^2 b \left(b-1\right) t^{b-2} s^{a+3}}{2\left(a+3\right)} + ... + \frac{\left(-1\right)^{b-2} b \left(b-1\right) t^2 s^{a+b-1}}{2\left(a+b-1\right)} + \frac{\left(-1\right)^{b-1} b t s^{a+b}}{a+b} + \frac{\left(-1\right)^b s^{a+b+1}}{a+b+1} |_0^t$$

$$=\frac{t^{a+b+1}}{a+1} + \frac{\left(-1\right)^1 b t^{a+b+1}}{a+2} + \frac{\left(-1\right)^2 b \left(b-1\right) t^{a+b+1}}{2\left(a+3\right)} + ... + \frac{\left(-1\right)^{b-2} b \left(b-1\right)} t^{a+b+1}}{2\left(a+b+1\right)} + \frac{\left(-1\right)^{b-1} b t^{a+b+1}}{a+b} + \frac{\left(-1\right)^b t^{a+b+1}}{a+b+1}$$

$$= t^{a+b+1} \left(\frac{\left(-1\right)^0 b!}{\left(a+1\right) 0! \left(b-0\right)!} + \frac{\left(-1\right)^1 b!}{\left(a+2\right) 1! \left(b-1\right)!} + \frac{\left(-1\right)^2 b!}{\left(a+3\right) 2! \left(b-2\right)!} + ... + \frac{\left(-1\right)^{b-2} b!}{\left(a+b-1\right) \left(b-2\right)! 2!} + \frac{\left(-1\right)^{b-1} b!}{\left(a+b\right) \left(b-1\right)! 1!} + \frac{\left(-1\right)^b b!}{\left(a+b+1\right) \left(b-0\right)! 0!} \right)$$

$$t^a \star t^b = t^{a+b+1} \sum_{n=0}^b \frac{\left(-1\right)^n b!}{\left(a+n+1\right) n! \left(b-n\right)!}$$

Does this seem right? And is it possible to generalize this to powers other than positive integers?

Last edited: May 25, 2010
2. May 27, 2010

### Char. Limit

Bumping, it's been two or three days.

I do want to know one thing: the formula I have only holds for positive integer b. Assuming everything I did holds to be true, can I simply change the factorials to gamma functions (changing the b to b-1 of course) and continue from there?